All categories

3 years ago

How does light travel?

Physics

2 answers:

Ivenika [448]3 years ago

7 0

Answer:

Explanation:

light can travel in a vacuum Anne the sped varies

devlian [24]3 years ago

6 0

The answer to this is C

You might be interested in

What is the toy's total energy at any point of its motion? Express your answer with the appropriate units.]

suter [353]

The complete question is as follows:

A 0.150-kg toy is undergoing SHM on the end of a horizontal spring with force constant k = 300 N/m. When the toy is 0.0120 m from its equilibrium position, it is observed to have a speed of 0.400 m/s.

Answer:

The correct answer is 0.034 J.

Explanation:

Given :

mass of the toy is m = 0.15 kg.

The force constant of restoring force k = 300 Nm⁻¹

When the position of the toy from the equilibrium is x = 0.012m, then the

speed of the toy vx = 0.4m s

The total mechanical energy in SHM is given by

E= 1/2 (mv²+ kx²) = 1/2 kA²

(here, m = mass of the object, vx = velocity, k = force constant

of restoring force, and A = amplitude of SHM.)

Hence by substituting the numerical values in equation 1, we get

$E= \frac{1}{2} (0.15* 0.4) + \frac{1}{2} (300* 0.012)$

= 0.034 J

Thus, the correct answer is 0.034 J.

6 0

3 years ago

A wire is oriented along the x-axis. It is connected to two batteries, and a conventional current of 2.4 A runs through the wire

Deffense [45]

Answer:

$\vec{F}=0.40176 N \hat{k}$

Explanation:

To calculate the force we need to use this equation

$\vec{F}=\int\limits^L_0 {i(\vec{dl}\times\vec{B})}$

where L is the total length of the wire

So in this case the small element of current is

$\vec{dl} = dx \hat{i}$

Because x is the direction of the current flow.

As is said in the problem B is such that

$\vec{B} = B \hat{j} = 0.62\hat{j} [ T]$

so to use the equation above we first calculate the following cross product:

$\vec{dl}\times\vec{B}=dx \hat{i}\times B \hat{j} = Bdx\hat{k}$

so the force:

$F = \int\limits^L_0 {i(\vec{dl}\times\vec{B})}=\int\limits^L_0{iBdx\hat{k}}$

So here we use the fact that B=0 in any point of the x axis that is not $x^{'}=0.27 [m]$ , that means that we only need to do the integration between a very short distant behind the point $x^{'}=0.27 [m]$ and a very short distant after that point, meaning:

$\vec{F}= \lim_{h \to 0}{\int\limits^{x^{'}+h}_{x^{'}-h}{iBdx\hat{k}} }$

so is the same as evaluating $iBx$ at $x=x^{'}$

that is:

$2,4 A * 0,62 T * 0,27 m \hat{k}$

$2,4 A * 0,62 (\frac{Kg}{A s^{2}}) * 0,27 m \hat{k}$

$2,4*0,62*2,7 ( \frac{ kgm }{ s^{2} } ) \hat{k}$

$\vec{F}=0.40176 N \hat{k}$

5 0

3 years ago

How to keep my family safe after a hurricane?

White raven [17]

Keep a first aid kit around
extra battery’s
extra water and food
have spare money set aside
be prepared for power outage
protect ur house with woof if necessary
keep surrounding yard clean of any big material

5 0

3 years ago

Read 2 more answers

When there are two light sources then there will be two shadows. Given below is a case where there are two light sources but one

velikii [3]

neither, a shadow cant be darker unless your closer to the surface your pointing the light at.

3 0

3 years ago

When making maps of the large-scale universe, astronomers estimate distances to the vast majority of galaxies by using:

Vesnalui [34]

Answer:

<em>The comoving distance and the proper distance scale</em>

Explanation:

The comoving distance scale removes the effects of the expansion of the universe, which leaves us with a distance that does not change in time due to the expansion of space (since space is constantly expanding). The comoving distance and proper distance are defined to be equal at the present time; therefore, the ratio of proper distance to comoving distance now is 1. The scale factor is sometimes not equal to 1. The distance between masses in the universe may change due to other, local factors like the motion of a galaxy within a cluster. Finally, we note that the expansion of the Universe results in the proper distance changing, but the comoving distance is unchanged by an expanding universe.

4 0

3 years ago